Perpendiculars on a line segment Two points A and B are given. Find the set of feet of the perpendiculars dropped from the point A onto all possivle straight lines passing through the point B.
 A: Let $O$ be the midpoint of $AB$ and let $AO=R$ . If $l$ is any line passing through $B$ and $C$ is the leg of the perpendicular of $A$ on $l$ then the triangle $ABC$ is right at $C$ and hence $CO=AO=BO$.
Thus $C$ is on the circle with center at $O$ and radius $R$. 
This shows that the locus is a subset of this circle. 
Moreover, if $C'$ is any point on this circle excepting $A$ and $B$, since $AB$ is a diameter then $<C'=90^0$ which shows that $C$ is a point on the locus.
If $C'=A$ then this is the case $l=AB$ and if $C'=B$ then this is the case $l \perp AB$.
This shows that the locus is the circle with diameter $AB$.
A: It seems a perpendicular from $A$ to a line through $B$ can be any length between $0$ and $|\overline{AB}|$, that is, if $L$ is the set of all possible lengths we have
$$L=\{l\mid0\le l\le|\overline{AB}|\}$$
The minimum occurs when the line is through $A$, and the maximum occurs when the line is perpendicular to the segment $\overline{AB}$.  All values in between occur as the line rotates about $B$.
